Kepler's Laws Quiz

Core fact. Kepler's first law (1609, Astronomia Nova): “The orbit of every planet is an ellipse with the Sun at one of the two foci.” For Earth satellites, the centre of the Earth occupies one focus; the second focus is empty. A circle is the special case e = 0.

Why the others are wrong:

• A — the older Platonic ideal; Brahe's Mars data specifically contradicted the circle-centred model — that contradiction is what forced Kepler to the ellipse.
• C — confuses the first law with the second; speed is not constant on an ellipse.
• D — states the third law (and misstates it: period squared is proportional to a cubed).

Core fact. Kepler's empirical “T² ∝ a³” became exact when Newton supplied the constant: T² = 4π²a³/μ, i.e. T = 2π√(a³/μ), with μ = GM⊙ = 3.986×10⁵ km³/s². This is the single most-used formula in satellite engineering.

Why the others are wrong:

• A — inverted structure (and would give units of s^-1, not seconds).
• B — Kepler's original proportionality is true only in units where the constant happens to be 1; with SI or km units you must include 4π²/μ.
• D — this is the circular velocity formula v = √(μ/r) with 2π attached — a unit-error distractor.

Exam tip. Always write a = R⊙ + h before substituting; the most common third-law error is feeding an altitude into the formula in place of a.

Core fact. Eccentricity measures deviation from circularity: e = c/a where c is the centre-to-focus distance. e = 0 ⇒ c = 0 ⇒ both foci coincide at the centre ⇒ a circle, with r_p = r_a = a.

Why the others are wrong:

• B — the escape (parabolic) case is e = 1, not 0.
• C — highly elliptical orbits have e approaching 1 (e.g., Molniya e ≈ 0.72).
• D — circular orbits are common: GEO comsats and GPS satellites are designed with e < 0.01.

Core fact. From the geometry of the ellipse, perigee and apogee radii (measured from the focus) are r_p = a(1−e) and r_a = a(1+e). Their sum is 2a (consistent: the major-axis length), and their ratio gives e = (r_a − r_p)/(r_a + r_p).

Why the others are wrong:

• A — this is the apogee radius, not the perigee — the classic swapped-formula trap.
• B — a√(1−e²) is the semi-minor axis b, a length measured from the ellipse centre, not a focus-radius.
• C — a/e is the distance from the centre to the directrix, not a perigee relation.

Sanity check. As e → 0, r_p → a (circle); as e → 1, r_p → 0 (degenerate) — both limits confirm D.

Core fact. Kepler's second law: equal areas swept in equal times. Angular momentum conservation (h = r²ω = r v_θ = constant) forces the transverse speed to be greatest where r is smallest — at perigee — and smallest at apogee. For the Molniya orbit this is a 6× speed swing (9.97 km/s vs. 1.62 km/s).

Why the others are wrong:

• A — true only for circles (e = 0); the question concerns elliptical orbits.
• C — exactly backwards.
• D — no special dynamics occur at the minor-axis crossings; speed there is intermediate between the perigee and apogee extremes.

Concept. Gravity acts along the radius vector (a central force), so its torque about Earth's centre is τ = r × F = 0. With zero torque, angular momentum h = r × mv is conserved: r²(dθ/dt) = constant. Multiplying by ½, the swept area per unit time dA/dt = h/2m is constant — exactly Kepler's second law. The second law is therefore angular-momentum conservation expressed geometrically.

Why the others fail:

• A — energy conservation is also true, but it governs the speed magnitude's variation via the vis-viva relation, not the equal-area geometry; the direct derivation of the second law is via h-conservation.
• B — reverses causality: the constant-area property holds for any central-force orbit; the ellipse itself comes from the inverse-square magnitude of the force.
• D — drag is a perturbation, absent from Keplerian mechanics, and above ~200 km it is negligible anyway.

Concept. Kepler's third law contains only a — eccentricity appears nowhere. Intuitively: the period is set by the orbit's energy ε = −μ/2a (same a ⇒ same energy), and a satellite on the ellipse oscillates in radius between r_p and r_a while a circular satellite stays at a constant r = a — both average out to the same a over one revolution. For a = 10,000 km: T = 2π√(10¹²/3.986×10⁵) ≈ 9,950 s ≈ 2 h 46 min for both.

Why the others fail:

• B/C — the ellipse's path is longer than the circle's circumference, but the elliptical satellite travels faster near perigee precisely so that the extra distance is covered in the same time — this is the second law compensating.
• D — the third law needs no eccentricity; a alone is sufficient.

Concept. Kepler's third law makes period a single-valued function of a. To hover over one longitude, the satellite must share Earth's spin rate: T = 86,164 s. Substituting: a = [μ(T/2π)²]^1/3 = 42,164 km — a unique solution. No other radius can give a 23 h 56 min period. Altitude h = a − R⊙ = 35,786 km follows.

Why the others fail:

• A — drag is negligible far below GEO; it does not “vanish” at any special altitude and plays no role in fixing the geostationary radius.
• B — solar radiation pressure (1 kW/m², μN/m² class) is a perturbation millions of times weaker than gravity; it never balances gravity at any radius.
• C — battery design is a consequence of eclipse geometry at GEO, not a determinant of the orbit radius.

Concept. By the second law, angular velocity ω = h/r² is smallest at maximum r — apogee. For Molniya (e ≈ 0.72), integrating the slow sweep near apogee shows ~8 of the 11 h 58 min period elapses within the high-altitude half of the orbit. This is a designed behaviour: the argument of perigee (ω ≈ 270°) places apogee over the northern hemisphere, so three satellites deliver continuous high-latitude service.

Why the others fail:

• A — the first law supplies the ellipse's shape but says nothing about time spent anywhere; dwell is a second-law phenomenon.
• C — the third law fixes the 12-h period but not the apportionment of time within the orbit.
• D — magnetic forces on the spacecraft are micro-Newton class; they cannot decelerate a 5-tonne satellite.

Concept. Kepler/Newton two-body motion assumes: (1) only two bodies, (2) spherically symmetric masses, (3) no other forces. Reality violates all three mildly: Earth's equatorial bulge (J2) twists the orbital plane (nodal regression) and rotates the apsides; the Moon and Sun tug (third body); residual atmosphere drags LEO satellites; sunlight pushes. These perturbations are small corrections (~10^-3 of the central acceleration), so the ellipse remains an excellent first approximation — which is why TLE-based tracking works.

Why the others fail:

• A — Newton did not disprove Kepler; he explained him — the laws are corollaries of gravitation plus F = ma.
• B — station-keeping burns are seconds-long, m/s-class events, not continuous thrust.
• D — there is no friction in space; eccentricity decays only through drag in the far upper atmosphere (a LEO effect).

Step-by-step:

Checks. Result lies in the LEO band (90–127 min) — consistent. Option A (85 min) is the period at sea-skimming altitude; C (118 min) results from using a = 6,578 km (subtracting 800 km instead of adding — a sign-error trap); D (144 min) corresponds to a ≈ 8,600 km (a = 7,800 km altitude error, i.e., confusing a with h twice).

Marking guide. Full marks: correct a = R⊙ + h, correct substitution, minutes conversion (divide by 60, not 3,600 twice).

Step-by-step:

Interpretation. e ≈ 0.72 identifies the classic Molniya HEO — and the 1,000/39,700 km altitude pair is its textbook signature. (a = 26,717 km comes out in passing; T = 2π√(a³/μ) ≈ 43,260 s ≈ 12 h 01 min, the half-sidereal-day Molniya period.)

Checks. Option A (0.32) results from dividing the altitude difference (38,700) by r_a alone; B (0.55) from dividing by 2a but with radii instead of (correct) 2a = 53,456 in numerator mishandling — i.e., (r_a−r_p)/r_a variants; C (0.64) from using h_a in place of r_a. The robust route is r_p, r_a first, e second.

Step-by-step (two independent routes):

Interpretation. This ~6× speed swing over one orbit is why HEO ground stations track rapidly through perigee and why link budgets vary enormously over the orbit.

Checks. Option A (1.7) is (1+e) alone; B (3.1) is √((1+e)/(1−e)) — the classic square-root slip; D (12.3) is (1+e)/(1−e)². The clean formula to memorise: v_p/v_a = (1+e)/(1−e).

Step-by-step:

Why the traps matter. Option B (42,164 km) is a itself — the radius from Earth's centre, not the altitude. The a-vs-h distinction is one of the most frequently examined concepts in this course: every third-law calculation uses a (centre-based), but every mission statement quotes h (surface-based).

Checks. C (48,542 km) adds R⊙ instead of subtracting; D (36,000 km) is a rounded “about 36,000 km” value — acceptable in conversation but not exact, and the question demands the exact figure.

Step-by-step (vis-viva):

Structure of the answer. The vis-viva equation covers all cases: circular (v = √(μ/r)), elliptical, and escape (v = √(2μ/r)). Note v_esc = √2 × v_circ = 1.414 × 7.67 — the √2 factor is worth memorising.

Checks. A (7.67 km/s) is the circular speed — forgetting the √2; B (9.11 km/s) results from v = √(3μ/2r) or a factor-of-√(3/2) error; D (15.34 km/s) is √2 × 10.84 — applying √2 twice.

Analysis. Decompose the comparison by law:

• Period (third law): T depends only on a. Both satellites: T = 2π√(26,578³/3.986×10⁵) = 43,080 s = 11 h 58 min. Equal.
• Energy: ε = −μ/2a — again a only. Equal.
• Motion (second law): the circular satellite (e = 0) keeps r = a and v = √(μ/a) = 3.87 km/s constant; the elliptical one sweeps between r_p = a(1−e) = 13,289 km and r_a = a(1+e) = 39,867 km, its speed oscillating between v_p = 6.2 km/s and v_a = 2.1 km/s.

Why the others fail: B/C — the second law exactly cancels path-length differences, as in Q7; D — perigee altitude is derivable (13,289 − 6,378 ≈ 6,911 km) but irrelevant to the period comparison; C — “circular efficiency” is not an energy concept; a alone sets ε.

Requirement-to-law mapping. “Dwell high over one hemisphere for hours” is a statement about time apportionment within the orbit — exactly what the second law governs (slow sweep near apogee). The third law only sets the total period; the first law only sets the shape.

Design synthesis. Combine all three laws: (1) first law — choose high e (≈ 0.7) so apogee is far (long dwell) and perigee low (cheap launch); (2) second law — the e ≈ 0.72 geometry yields ~8 h of the 12 h near apogee; (3) third law — choose a = 26,578 km for the half-sidereal-day 11 h 58 min period, and set argument of perigee ω ≈ 270° with i = 63.4° so apogee parks over the northern hemisphere (frozen apsides). Three satellites phased 8 h apart give continuous coverage.

Why the others fail: A — GEO gives continuous visibility but at fixed low elevation for high latitudes and no “dwell” phenomenon; B — a circular orbit has no apogee at all; D — orbital speed cannot be “artificially reduced” — that would change the orbit itself (speed is set by r and a via vis-viva).

Step-by-step:

Interpretation. The 3/2 power is the engine of orbit design: a modest 5.9× increase in radius buys a 14× longer period, which is why a single GEO satellite can match Earth's rotation while LEO completes 14–15 revolutions a day.

Checks. B uses a² (34.5×) — treating T ∝ a²; C uses the linear ratio (5.9×) — forgetting the power entirely; D invokes satellite mass, which cancels from the two-body equation (Galileo's principle: all bodies fall equally) — mass never appears in Kepler's laws.

Step-by-step:

Transfer physics. The Hohmann logic uses all three laws: (1) the first law defines the transfer ellipse touching LEO-class perigee and GEO apogee; (2) the second law says the satellite crawls near apogee (v ≈ 1.6 km/s there) — the ideal place to fire the apogee-kick motor, since a given burn changes the specific energy efficiently and the geometry stays fixed long enough to aim; (3) the third law sets the coast time at half the transfer period, ~5.3 h–10.5 h depending on phasing (in practice ~5.3 h for half an orbit — the full period is 10.5 h, and the coast from perigee to apogee is half a revolution ≈ 5.3 h; note the option quotes the full period scale for context).

Why the others fail: B — a = 42,164 km with e = 0 describes the final GEO orbit, not the transfer; C — perigee burns are efficient for raising apogee (Oberth effect), but the quoted a = 18,532 km is wrong and the answer misidentifies where circularisation happens; D — Kepler's laws apply to every orbit, transfer or operational, equally.

Element-by-element analysis:

• a = 42,164 km ⇒ T = 86,164 s (third law) ⇒ the satellite is geosynchronous — its mean motion matches Earth's spin.
• e = 0.0005 → r oscillates by ±a·e ≈ ±21 km; by the second law the satellite speeds and slows slightly, drifting east–west about the nominal longitude.
• i = 0.4° → the sub-satellite point also moves north–south by ±0.4° of latitude (≈ ±45 km) each day.

Combined, the ground track is a tiny analemma (figure-8) centred on the assigned longitude. Antennas with ~0.1° beamwidths must either track or accept a small pointing loss; this is why operators perform periodic inclination-control (north–south station-keeping, ~50 m/s of ΔV per year) to keep i near zero — or deliberately allow i to grow to save propellant, trading tracking burden for lifetime.

Why the others fail: A — e and i are small but physically consequential, and station-keeping exists precisely because they grow; C — Molniya has e ≈ 0.72 and a = 26,578 km, three orders of magnitude different in eccentricity; D — e and i can never be held at exactly zero (perturbations guarantee drift); operationally “geostationary” means within a deadband.